Question

The integers $r$ and $k$ are randomly selected, where $-3 < r < 6$ and $1 < k < 8$. what is the probability that the division $r \div k$ is an integer value? express your answer as a common fraction.

Answer 1

Denote by $R,K$ the random variables representing the integer values $r,k$, respectively. Then $R\sim\mathrm{Unif}(-2,5)$ and $K\sim\mathrm{Unif}(2,7)$, where $\mathrm{Unif}(a,b)$ denotes the discrete uniform distribution over the interval $[a,b]$. So $R$ and $K$ have probability mass functions

$p_R(r)=\begin{cases}\dfrac18&\text{for }r\in\{-2,-1,\ldots,5\}\\\\0&\text{otherwise}\end{cases}$

$p_K(k)=\begin{cases}\dfrac16&\text{for }k\in\{2,3,\ldots7\}\\\\0&\text{otherwise}\end{cases}$

We want to find $P\left(\dfrac RK\equiv0\pmod n\right)$, where $n$ is any integer.

We have six possible choices for $K$:

(i) if $K=2$, then $\dfrac RK$ is an integer when $R=\pm2,0,4$;

(ii) if $K=3$, then $\dfrac RK$ is an integer when $R=0,3$;

(iii) if $K=4$, then $\dfrac RK$ is an integer when $R=0,4$;

(iv) if $K=5$, then $\dfrac RK$ is an integer when $R=0,5$;

(v) if $K=6$ or $K=7$, then $\dfrac RK$ is an integer only when $R=0$ in both cases.

If the selection of $R,K$ are made independently, then the joint distribution is the product of the marginal distribution, i.e.

$p_{R,K}(r,k)=p_R(r)\cdot p_K(k)=\begin{cases}\dfrac1{48}&\text{for }(r,k)\in[-2,5]\times[2,7]\\\\0&\text{otherwise}\end{cases}$

That is, there are 48 possible events in the sample space. We counted 12 possible outcomes in which $\dfrac RK$ is an integer, so the probability of this happening is $\dfrac{12}{48}=\dfrac14$.